Question

Elemental s reacts with o2 to form so3 according to the reaction 2s+3o2→2so3a.) how many o2 molecules are needed to react with 3.50 g of s?b.) what is the theoretical yield of so3 produced by the quantities described in part a?

Answer 1

Answer :

The number of molecules of $O_2$ are, $9.88\times 10^{22}$

The theoretical yield of $SO_3$ is, 8.72 grams.

Explanation :

The given balanced chemical reaction is:

$2S+3O_2\rightarrow 2SO_3$

First we have to calculate the moles of S.

$\text{Moles of }S=\frac{\text{Mass of }S}{\text{Molar mass of }S}$

Molar mass of S = 32 g/mole

$\text{Moles of }S=\frac{3.50g}{32g/mol}=0.109mole$

Now we have to calculate the moles of $O_2$

From the balanced chemical reaction, we conclude that:

As, 2 mole of $S$ react with 3 mole of $O_2$

So, 0.109 moles of $S$ react with $\frac{0.109}{2}\times 3=0.164$ moles of $O_2$

Now we have to calculate the number of molecules of $O_2$

As, 1 mole of $O_2$ contains $6.022\times 10^{23}$ number of $O_2$ molecules.

So, 0.164 mole of $O_2$ contains $0.164\times 6.022\times 10^{23}=9.88\times 10^{22}$ number of $O_2$ molecules.

Thus, the number of molecules of $O_2$ are, $9.88\times 10^{22}$

Now we have to calculate the moles of $SO_3$

From the reaction, we conclude that

As, 2 mole of $S$ react to give 2 mole of $SO_3$

So, 0.109 moles of $S$ react to give 0.109 moles of $SO_3$

Now we have to calculate the mass of $SO_3$

$\text{ Mass of }SO_3=\text{ Moles of }SO_3\times \text{ Molar mass of }SO_3$

Molar mass of $SO_3$ = 80 g/mol

$\text{ Mass of }SO_3=(0.109moles)\times (80g/mole)=8.72g$

Thus, the theoretical yield of $SO_3$ is, 8.72 grams.